Understanding Lens Experiment Results: Focal Length and Graph Interpretation

[shakerist]

Well at school we are doing an experiment to find out the factors which affect the distance between the lens and the image.

Well I did the convex lens experiement and from my results i plotted a graph which is y vs x
The distance between the lightbulb and lens is x and the distance between the lens and image is Y ( is this called the focal length). Looking around the internet everone has it V vs U. Is V my distance y and U my distance X ?

Another question is that my 1/y vs 1/x graph is wrong because everyone's graphs I've seen on the net are from the top left going down while mine starts from the bottom left and goes up.

The file attached contains images which show the shape of the graph.

Thanks

 

[Delphi51]

Welcome to PF, shakerist.
Those letters sure are confusing. I'm used to Di for the distance from center of lens to image, Do for distance from lens to object (light). I'm so old we used candles for the light source. The formula in those days was 1/f = 1/Di + 1/Do, where f is the focal length. This conflicts with your comment that the distance from the lens to the image is the focal length; better check that out. Also, my formula appears more complicated than your simple inverse variation graph. The graph should be y vs x (Di vs Do) because you are asked how Do affects Di.

It would be interesting to see some of your measurements. Also, did you get the focal length by holding the lens up to a distance light source (room lights several meters away) and measuring the distance from lens to image on a bit of paper?

 

[shakerist]

Hi,
Well I did it Di vs Do and it is a straight line from the top left going down to the bottom right. I have a gradient of -1 and from my understanding f= gradient so my f is -1. 1/-1 is -1 and when i add my Di and Do i don't get -1 instead i get 0.09 ... which is positive not negative. So what am i doing wrong ?

 

[shakerist]

my y intercept is 0.0945 and my x intercept is also 0.0945 if that helps. Also the ligh source was a small lightbulb. We changed the distance of the lightbulb and took a reading of Di. I am not giving a lot of detail as it might go down as cheating

 

[shakerist]

Anyone help please

 

[Redbelly98]

Can you post your data, and show a calculation of what you are trying to do? We can't tell what is going wrong unless you show us what you did.

 

[shakerist]

Here are my results, and i am trying to show that my results are accurate and also trying to show 1/f = 1/Di + 1/Do from my graphs, sorry for being such an idiot I am a gcse physics student and i have no idea what I am doing, one thing for sure i won't be picking it for A levels :P

 

[Redbelly98]

Hi,
Well I did it Di vs Do and it is a straight line from the top left going down to the bottom right. I have a gradient of -1 and from my understanding f= gradient so my f is -1. 1/-1 is -1 and when i add my Di and Do i don't get -1 instead i get 0.09 ... which is positive not negative. So what am i doing wrong ?

Okay, I have thought about this some more. No, the gradient (what we normally call the slope in the USA) is not f. Try working with the equation:

1/Do + 1/Di = 1/f​

or

x + y = 1/f​

where x=1/Do and y=1/Di. Do two things:

1. Solve the equation for y (make y the subject). Now do you see how 1/f relates to the graph of x and y?

2. Plot y vs. x. You have the y values in your data, you'll need to make another column for x values (=1/Do).

As another tip, when Do and Di are equal, they are each equal to 2f as well. You can find this value from your data, since Do and Di are nearly equal for one of your measurements. That is a way to check what you get for f. But if your instructions are to find it graphically, then you should do it that way as well.

 

[shakerist]

so my 1/f is 0.0945, y= 0.0945 - 0.06
so y is = 0.0345, on my graph when x = to 0.06 y = 0.0345 so i guess I've finally managed it.

Thanks guys